Iron can be formed according to the following reaction: Fe3O4(s) + 4H2(g) ⇌ 3Fe(s) + 4H2O(g)...

Iron can be formed according to the following reaction: Fe3O4(s) + 4H2(g) ⇌ 3Fe(s) + 4H2O(g) Kc = 0.00178 at 25 oC In a 1.00 L container is 2.13 grams of Fe3O4, 1.27 grams of Fe, and 0.9377 grams of H2. If the reaction is at equilibrium, how many grams of H2O are in the container?

Solutions

Expert Solution

Fe3O4(s) + 4H2(g) ⇌ 3Fe(s) + 4H2O(g)

no of moles of H2 = W/G.M.Wt = 0.9377/2 =0.46885 moles

Fe3O4(s) + 4H2(g) ⇌ 3Fe(s) + 4H2O(g)

initial - 0.46885 - 0

Change - -4x 4x

at equilibrium - 0.46885-4x 4x

Kc = [Fe]³[H2O]⁴/[Fe3O4][H2]⁴

Fe and Fe3O4 are solids conc of Fe and Fe3O4 are one

Kc = [H2O]⁴/[H2]⁴

0.00178 = (4x)⁴/(0.46885-x )⁴

(0.00178)^1/4 = 4x/0.46885-x

0.2054 = 4x/0.46885-x

0.0963 - 0.2054x = 4x

0.0963-0.2054x-4x =0

4.2054x = 0.0963

x=0.0963/4.2054 = 0.0228

[H2O] = 4X

= 4*0.0228 = 0.0912 moles

mass of water = no of moles of water*molar mass

= 0.0912*18 = 1.6416gm >>>> answer

queen_honey_blossom answered 1 year ago

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